Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
struct node {
  int num[12], end;
  int& operator[](const int& pos) { return num[pos]; }
  void push_back(int a) { num[++end - 1] = a; }
  int size() { return end; }
  void clear() {
    memset(num, 0, sizeof(num));
    end = 0;
  }
  void Sort() { sort(num, num + end); }
} b[maxn], d[maxn][21];
int n, m, q, ai, bi, k, cnt, head[maxn], p[maxn][21], dep[maxn], temp[maxn],
    tempsize;
vector<int> ba[maxn];
struct Edge {
  int v, nxt;
} e[maxn * 2];
void add(int u, int v) {
  e[++cnt].v = v;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
void dfs(int u, int fa) {
  dep[u] = dep[fa] + 1;
  p[u][0] = fa;
  memset(temp, 0, sizeof(int) * tempsize);
  tempsize = 0;
  for (int i = 0; i < b[fa].size(); i++) temp[++tempsize - 1] = b[fa][i];
  sort(temp, temp + tempsize);
  for (int i = 0; i < min(10, tempsize); i++) d[u][0].push_back(temp[i]);
  for (int i = 1; i <= 17; i++) {
    p[u][i] = p[p[u][i - 1]][i - 1];
    memset(temp, 0, sizeof(int) * tempsize);
    tempsize = 0;
    for (int j = 0; j < d[u][i - 1].size(); j++)
      temp[++tempsize - 1] = d[u][i - 1][j];
    for (int j = 0; j < d[p[u][i - 1]][i - 1].size(); j++)
      temp[++tempsize - 1] = d[p[u][i - 1]][i - 1][j];
    sort(temp, temp + tempsize);
    for (int j = 0; j < min(10, tempsize); j++) d[u][i].push_back(temp[j]);
  }
  for (int i = head[u]; i != -1; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa) continue;
    dfs(v, u);
  }
}
int lca(int a, int b) {
  if (dep[b] > dep[a]) swap(a, b);
  if (dep[b] != dep[a]) {
    for (int i = 17; i >= 0; i--)
      if (dep[p[a][i]] < dep[b])
        continue;
      else
        a = p[a][i];
  }
  if (a == b) return a;
  for (int i = 17; i >= 0; i--) {
    if (p[a][i] == p[b][i]) continue;
    a = p[a][i];
    b = p[b][i];
  }
  return p[a][0];
}
int main() {
  while (scanf("%d%d%d", &n, &m, &q) == 3) {
    memset(head, -1, sizeof(head));
    cnt = 0;
    for (int i = 1; i < n; i++) {
      scanf("%d%d", &ai, &bi);
      add(ai, bi);
      add(bi, ai);
    }
    for (int i = 1; i <= m; i++) {
      scanf("%d", &ai);
      ba[ai].push_back(i);
    }
    for (int i = 1; i <= n; i++) {
      sort(ba[i].begin(), ba[i].end());
      for (int j = 0; j < min(10, int(ba[i].size())); j++)
        b[i].push_back(ba[i][j]);
    }
    dfs(1, 0);
    for (int h = 1; h <= q; h++) {
      scanf("%d%d%d", &ai, &bi, &k);
      int LCA = lca(ai, bi);
      memset(temp, 0, sizeof(int) * tempsize);
      tempsize = 0;
      for (int i = 0; i < b[ai].size(); i++) temp[++tempsize - 1] = b[ai][i];
      for (int i = 0; i < b[bi].size(); i++) temp[++tempsize - 1] = b[bi][i];
      for (int i = 17; i >= 0; i--)
        if (dep[p[ai][i]] >= dep[LCA]) {
          for (int j = 0; j < d[ai][i].size(); j++)
            temp[++tempsize - 1] = d[ai][i][j];
          ai = p[ai][i];
        }
      for (int i = 17; i >= 0; i--)
        if (dep[p[bi][i]] >= dep[LCA]) {
          for (int j = 0; j < d[bi][i].size(); j++)
            temp[++tempsize - 1] = d[bi][i][j];
          bi = p[bi][i];
        }
      sort(temp, temp + tempsize);
      int SIZE = tempsize, ok = 0;
      for (int i = 1; i < tempsize; i++)
        if (temp[i] == temp[i - 1]) SIZE--;
      SIZE = min(SIZE, k);
      printf("%d ", SIZE);
      if (SIZE != 0) printf("%d ", temp[0]), ok++;
      for (int i = 1; i < tempsize; i++)
        if (temp[i] != temp[i - 1] && ok < SIZE) printf("%d ", temp[i]), ok++;
      printf("\n");
    }
  }
}